Preliminary

Jordan Matrix

Jordan Normal Form Theorem

$A\in {\mathbb{C}}^{n\times n}$, $P_A(\lambda )=\prod _{i=1}^s(\lambda -{\lambda }_i{)}^{n_i}$, then it is similar to Jordan normal form $diag(J_1,J_2,\cdots ,J_s)$, $J_i\in {\mathbb{n}}_{\mathbb{i}}\times {\mathbb{n}}_{\mathbb{i}}$, where $n_i$ is algebraic multiplicity of $A$,

$$J_i=diag(J_{i_1},J_{i_2},\cdots ,J_{i_{m_i}}),m_i\text{ is the geometric multiplicity of }A$$

$$J_{ij}=J_{l_{ij}}({\lambda }_i)=\left[\begin{array}{cc}{\lambda }_i& 1\\ & {\lambda }_i& \ddots \\ & & \ddots & 1\\ & & & {\lambda }_i\end{array}\right]\in {\mathbb{C}}^{l_{ij}\times l_{ij}}$$

And the Jordan normal form is unique if we do not consider the sequence of small blocks.

Minimal polynomial

Matrix $A$ has a minimal polynomial

$$d_A(\lambda )=d_J(\lambda )=\prod _{i=1}^s(\lambda -{\lambda }_i{)}^{k_i}$$

where

$$k_i=\underset{1\le j\le m_i}{max}\{l_{ij}\}$$

$m_i$ determines how much blocks compose the whole Jordan normal form. For a Jordan normal form $J_i$, with $n_i$ algebraic multiplicity of ${\lambda }_i$, has $m_i$ number of Jordan small blocks. We can easily see that the prime diagonal has $n_i-m_i$ number of $1$, which means rank$(J-{\lambda }_{i}I)=n_i-m_i$. Then to confirm the composition of these small blocks, we have to write $n_i-m_i$ as a summation of $m_i$ number non-negative integers, each of which corresponds to a Jordan matrix.

Not complicatedly speaking, we have to calculate the power of $J_i-{\lambda }_i$ to get the result.

Denote ${\delta }_p^i$ as the number of $p$th Jordan small blocks corresponding to eigenvalue ${\lambda }_i$, where $p=1,2\cdots ,n_i$, so

$$\begin{array}{r}\sum _{p=1}^{n_i}{\delta }_p^i=m_i,\text{total number of blocks}\\ \sum _{p=1}^{n_i}p{\delta }_p^i=n_i\text{total rank of Jordan}\end{array}$$

Introduce

$$r_k^i=r(A-{\lambda }_{i}I{)}^k$$

so

$$\begin{array}{rl}{r}_1^i& =r(A-{\lambda }_{i}I)\\ & =r(J-{\lambda }_{i}I)\\ & =n-n_i+r(J_i-{\lambda }_{i}I)\\ & =n-n_i+\sum _{p=1}^{n_i}(p-1){\delta }_p^i\text{1 order block becomes 0 when substraction}\end{array}$$

So after $k$ times power

$$r_k^i=n-n_i+\sum _{p=k}^{n_i}(p-k){\delta }_p^i\text{Why?}$$

and

$$\begin{array}{rl}{r}_{k-1}^i& =n-n_i+\sum _{p=k-1}^{n_i}(p-k+1){\delta }_p^i\\ & =n-n_i+\sum _{p=k}^{n_i}(p-k+1){\delta }_p^i\end{array}$$

If we define

$$d_k^i:=r_{k-1}^i-r_k^i=\sum _{p=k}^{n_i}{\delta }_p^i$$

and

$$d_{k+1}^i=\sum _{p=k+1}^{n_i}{\delta }_p^i$$

Substract the aboce two equaiton, we get

$${\delta }_p^i=d_k^i-d_{k+1}^i$$

So if a Jordan small block has ${\delta }_p^i$ number, then the rank decline from power $p-1$ to power $p$ of $(A-{\lambda }_{i}I)$ would display.